find the equation the directrix of given ellipse `x^(2) + 16y^(2) = 16 `

Answer

To find the equation of the directrices of the given ellipse \( x^2 + 16y^2 = 16 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation can be rewritten in standard form by dividing all terms by 16:

\[ \frac{x^2}{16} + \frac{y^2}{1} = 1 \]

This shows that \( a^2 = 16 \) and \( b^2 = 1 \).

### Step 2: Identify \( a \) and \( b \)
From the standard form, we find:

\[ a = \sqrt{16} = 4 \quad \text{and} \quad b = \sqrt{1} = 1 \]

### Step 3: Calculate the eccentricity \( e \)
The eccentricity \( e \) of the ellipse is given by the formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting the values of \( a^2 \) and \( b^2 \):

\[ e = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{16 - 1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]

### Step 4: Find the equations of the directrices
The equations of the directrices for an ellipse are given by:

\[ x = \pm \frac{a}{e} \]

Substituting the values of \( a \) and \( e \):

\[ x = \pm \frac{4}{\frac{\sqrt{15}}{4}} = \pm \frac{4 \times 4}{\sqrt{15}} = \pm \frac{16}{\sqrt{15}} \]

### Final Answer
Thus, the equations of the directrices are:

\[ x = \frac{16}{\sqrt{15}} \quad \text{and} \quad x = -\frac{16}{\sqrt{15}} \]